logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
0 votes

The decomposition of $N_2O_5$ is a first order reaction represented by $N_2O_5\rightarrow N_2O_4+\large\frac{1}{2}$$O_2$.After 15minutes the volume of $O_2$ produced is 9ml & at the end of the reaction 35ml.The rate constant is equal to

$\begin{array}{1 1}(a)\;\large\frac{1}{15}\normalsize ln\large\frac{44}{35}&(b)\;\large\frac{1}{15}\normalsize ln \large\frac{35}{44}\\(c)\;\large\frac{1}{15}\normalsize ln\large\frac{44}{26}&(d)\;\large\frac{1}{15}\normalsize ln\large\frac{35}{26}\end{array}$

Can you answer this question?
 
 

1 Answer

0 votes
Here $k=\large\frac{1}{t}$$ln\bigg(\large\frac{V_{\infty}-V_0}{V_{\infty}-V_t}\bigg)$
$V_{\infty}$-Volume of $O_2$ at end of reaction.
$V_0$-Volume of $O_2$ at t=0
$V_t$-Volume of $O_2$ at t=t
t=15minutes
$V_{\infty}-V_0=35ml$
$V_{\infty}-V_t=(35-9)ml=26ml$
$\therefore k=\large\frac{1}{15}$$ln\large\frac{35}{26}$
Hence (d) is the correct answer.
answered Dec 16, 2013 by sreemathi.v
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...